Question

A mass M is suspended by a rope from a rigid support at A as shown in figure. Another rope is tied at the end B, and it is pulled horizontally with a force F. If the rope AB makes an angle $\theta$ with the vertical, then the tension in the string AB is?

• A. F $\sin \theta$
• B. F$/\sin \theta$
• C. F$\cos \theta$
• D. F$/\cos \theta$

Answer 1

The correct option is B F$/\sin \theta$

We know that if three forces $F_1,F_2,F_3$ are acting at a point and angle opposite to them are $\alpha ,\beta ,\gamma$ and are at equilibrium.

Then $\frac{F_1}{\alpha }=\frac{F_2}{\beta }=\frac{F_3}{\gamma }$

Here,

$\frac{F}{\sin ({180}^o-\theta )}=\frac{T}{\sin {90}^o}=\frac{Mg}{\sin ({90}^o+\theta )}$

Using first two terms-

$\frac{F}{\sin \theta }=\frac{T}{1}$

$⟹T=\frac{F}{\sin \theta }$

Answer-(B)